In Case 1, \[x^2-3\sqrt{2}x+4=(x-\sqrt{2})(x-2\sqrt{2})=0,\] and so we have that \(x=\sqrt{2}\) and \(x=2\sqrt{2}\) are solutions. However, when \(x=\sqrt{2}\), we end up with \(0^0\).

The quantity \(0^0\) is a bit problematic. For some mathematicians there are good reasons to define \(0^0\) to be \(1\), whereas other mathematicians prefer to say that \(0^0\) is undefined.

So we may choose to discard \(x=\sqrt{2}\) as a solution.

In Case 2, we have that \[x^2-1=(x+1)(x-1)=0,\] so \(x=1\) and \(x=-1\) are also solutions.

Solving Case 3, we get \[x^2-3=(x+\sqrt{3})(x-\sqrt{3})=0,\] and so \(x=\pm\sqrt{3}\) are potential solutions. However, we have that \[\sqrt{3}^2\pm3\sqrt{2}\sqrt{3}+4=7\pm3\sqrt{6}\] is not divisible by two (it isn’t even an integer!) so \(x=\pm\sqrt{3}\) are not solutions.

So the values of \(x\) satisfying the equation are \[x=-1, 1, 2\sqrt{2}.\]